Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer and have also been studying schematics for
some historical analog computers. An integrator is modeled with an
inverting op amp with a feedback capacitor.

In the circle test, a simple loop is constructed modeling the
differential equation y'' = -y. This is implemented with

Integrator A -> Integrator B -> inverter -> loop back to input of
Integrator A.

In ideal integration, the function y — which is the output of Integrator
B — would be a steady sinusoidal wave that does not increase or decrease
in amplitude. However, in practice the wave will either grow or decay in amplitude, once the integrators are started. I've found that, with
smaller capacity feedback capacitors, the wave tends to grow, and with
larger ones, it tends to decay. One can see this as a growing or
shrinking circle on an oscilloscope, by feeding the outputs from both integrators into an XY scope display.

Could somebody please explain why this happens? I'm not grasping the
basic cause of this. I've been trying to read up about op amp
(in)stability, like in amplifiers and voltage followers, but I'm not
seeing if/how there is a connection between that and what is going on
here.

--
Christopher Howard

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard <[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer and have also been studying schematics for
some historical analog computers. An integrator is modeled with an
inverting op amp with a feedback capacitor.

In the circle test, a simple loop is constructed modeling the
differential equation y'' = -y. This is implemented with

Integrator A -> Integrator B -> inverter -> loop back to input of
Integrator A.

In ideal integration, the function y � which is the output of Integrator
B � would be a steady sinusoidal wave that does not increase or decrease
in amplitude. However, in practice the wave will either grow or decay in >amplitude, once the integrators are started. I've found that, with
smaller capacity feedback capacitors, the wave tends to grow, and with
larger ones, it tends to decay. One can see this as a growing or
shrinking circle on an oscilloscope, by feeding the outputs from both >integrators into an XY scope display.

Could somebody please explain why this happens? I'm not grasping the
basic cause of this. I've been trying to read up about op amp
(in)stability, like in amplifiers and voltage followers, but I'm not
seeing if/how there is a connection between that and what is going on
here.

Don't blame the poor opamps. They can't do y'' = -y perfectly. A
little extra phase shift always sneaks in.

At the peak frequency, if the loop gain is 1.001, the oscillation will
grow exponentially. If gain is 0.9999, it will decay exponentially.

A stable oscillator needs to be managed to keep amplitude constant.
That requires a nonlinear limiter of some sort.

There was a big (and fairly silly) thread here about that, a couple
months agot.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard ><[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer and have also been studying schematics for
some historical analog computers. An integrator is modeled with an >>inverting op amp with a feedback capacitor.

In the circle test, a simple loop is constructed modeling the
differential equation y'' = -y. This is implemented with

Integrator A -> Integrator B -> inverter -> loop back to input of >>Integrator A.

In ideal integration, the function y � which is the output of Integrator
B � would be a steady sinusoidal wave that does not increase or decrease
in amplitude. However, in practice the wave will either grow or decay in >>amplitude, once the integrators are started. I've found that, with
smaller capacity feedback capacitors, the wave tends to grow, and with >>larger ones, it tends to decay. One can see this as a growing or
shrinking circle on an oscilloscope, by feeding the outputs from both >>integrators into an XY scope display.

Could somebody please explain why this happens? I'm not grasping the
basic cause of this. I've been trying to read up about op amp >>(in)stability, like in amplifiers and voltage followers, but I'm not
seeing if/how there is a connection between that and what is going on
here.

Don't blame the poor opamps. They can't do y'' = -y perfectly. A
little extra phase shift always sneaks in.

At the peak frequency, if the loop gain is 1.001, the oscillation will
grow exponentially. If gain is 0.9999, it will decay exponentially.

A stable oscillator needs to be managed to keep amplitude constant.
That requires a nonlinear limiter of some sort.

There was a big (and fairly silly) thread here about that, a couple
months agot.

For which I was originally responsible for (before BS and Edward took
it to a whole new level).

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge oscillator's principle of operation and how they overcame this problem
to stabilise the loop in the most simple and elegant way.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 20/07/2025 5:58 pm, Cursitor Doom wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard
<[email protected]> wrote:

<snip>

For which I was originally responsible for (before BS and Edward took
it to a whole new level).

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge oscillator's principle of operation and how they overcame this problem
to stabilise the loop in the most simple and elegant way.

Cursitor Doom has delusions of competence.

There are a variety of ways of stabilising a Wein Bridge, and none of
them are all that elegant. What the thread did manage to establish was
that there were a couple of ways of getting the harmonic content of the nominally sinewave output close to 150dB below the fundamental nanovolts
in a notionally 1V peak-to-peak signal.

John May was the most influential contributor, mostly because he has
actually built examples of that kind of circuit. Edward Rawde was the
most persistent.

--
Bill Sloman, Sydney

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Sun, 20 Jul 2025 08:58:12 +0100, Cursitor Doom <[email protected]>
wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard >><[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer and have also been studying schematics for
some historical analog computers. An integrator is modeled with an >>>inverting op amp with a feedback capacitor.

In the circle test, a simple loop is constructed modeling the >>>differential equation y'' = -y. This is implemented with

Integrator A -> Integrator B -> inverter -> loop back to input of >>>Integrator A.

In ideal integration, the function y � which is the output of Integrator >>>B � would be a steady sinusoidal wave that does not increase or decrease >>>in amplitude. However, in practice the wave will either grow or decay in >>>amplitude, once the integrators are started. I've found that, with >>>smaller capacity feedback capacitors, the wave tends to grow, and with >>>larger ones, it tends to decay. One can see this as a growing or >>>shrinking circle on an oscilloscope, by feeding the outputs from both >>>integrators into an XY scope display.

Could somebody please explain why this happens? I'm not grasping the >>>basic cause of this. I've been trying to read up about op amp >>>(in)stability, like in amplifiers and voltage followers, but I'm not >>>seeing if/how there is a connection between that and what is going on >>>here.

Don't blame the poor opamps. They can't do y'' = -y perfectly. A
little extra phase shift always sneaks in.

At the peak frequency, if the loop gain is 1.001, the oscillation will
grow exponentially. If gain is 0.9999, it will decay exponentially.

A stable oscillator needs to be managed to keep amplitude constant.
That requires a nonlinear limiter of some sort.

There was a big (and fairly silly) thread here about that, a couple
months agot.

For which I was originally responsible for (before BS and Edward took
it to a whole new level).

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge >oscillator's principle of operation and how they overcame this problem
to stabilise the loop in the most simple and elegant way.

An simple opamp double-integrator oscilllator can be made to run very
close to 1.000 loop gain, so needs very little nonlinear gain tweaking
to be amplitude stable. A Wein bridge needs much more tweaking hence
inherits a lot of distortion from the gain control mechanism.

The thing that (d)evolved in that other thread became hilarious.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

"Bill Sloman" <[email protected]> wrote in message news:105ihmc$3ahv8$[email protected]...

On 20/07/2025 5:58 pm, Cursitor Doom wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard
<[email protected]> wrote:

<snip>

For which I was originally responsible for (before BS and Edward took
it to a whole new level).

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge
oscillator's principle of operation and how they overcame this problem
to stabilise the loop in the most simple and elegant way.

Cursitor Doom has delusions of competence.

There are a variety of ways of stabilising a Wein Bridge, and none of them are all that elegant. What the thread did manage to
establish was that there were a couple of ways of getting the harmonic content of the nominally sinewave output close to 150dB
below the fundamental nanovolts in a notionally 1V peak-to-peak signal.

John May was the most influential contributor, mostly because he has actually built examples of that kind of circuit. Edward Rawde
was the most persistent.

I'm still persisting but I don't currently have time to find out why a circuit which stabilizes and produces low distortion in
LTSpice just hits the power rail when the same circuit is run in QSpice.

--
Bill Sloman, Sydney

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 21/07/2025 1:24 am, john larkin wrote:

On Sun, 20 Jul 2025 08:58:12 +0100, Cursitor Doom <[email protected]>
wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard
<[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer and have also been studying schematics for
some historical analog computers. An integrator is modeled with an
inverting op amp with a feedback capacitor.

In the circle test, a simple loop is constructed modeling the
differential equation y'' = -y. This is implemented with

Integrator A -> Integrator B -> inverter -> loop back to input of
Integrator A.

In ideal integration, the function y — which is the output of Integrator >>>> B — would be a steady sinusoidal wave that does not increase or decrease >>>> in amplitude. However, in practice the wave will either grow or decay in >>>> amplitude, once the integrators are started. I've found that, with
smaller capacity feedback capacitors, the wave tends to grow, and with >>>> larger ones, it tends to decay. One can see this as a growing or
shrinking circle on an oscilloscope, by feeding the outputs from both
integrators into an XY scope display.

Could somebody please explain why this happens? I'm not grasping the
basic cause of this. I've been trying to read up about op amp
(in)stability, like in amplifiers and voltage followers, but I'm not
seeing if/how there is a connection between that and what is going on
here.

Don't blame the poor opamps. They can't do y'' = -y perfectly. A
little extra phase shift always sneaks in.

At the peak frequency, if the loop gain is 1.001, the oscillation will
grow exponentially. If gain is 0.9999, it will decay exponentially.

A stable oscillator needs to be managed to keep amplitude constant.
That requires a nonlinear limiter of some sort.

There was a big (and fairly silly) thread here about that, a couple
months agot.

For which I was originally responsible for (before BS and Edward took
it to a whole new level).

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge
oscillator's principle of operation and how they overcame this problem
to stabilise the loop in the most simple and elegant way.

An simple opamp double-integrator oscilllator can be made to run very
close to 1.000 loop gain, so needs very little nonlinear gain tweaking
to be amplitude stable. A Wein bridge needs much more tweaking hence
inherits a lot of distortion from the gain control mechanism.

Or so John Larkin thinks. He won't be able to explain why he thinks that
the Wein Bridge needs more tweaking, or a least not in terms that make
much sense. The problem with both is passive component tolerances and
drifts. Precision resistors are available with much tighter tolerances
and lower drifts than capacitors, but either sort of oscillator always
needs both.

The thing that (d)evolved in that other thread became hilarious.

John Larkin doesn't seem to have been able to follow the discussion, so
- to preserve his self-esteem - he decided it was hilarious.

--
Bill sloman, Sydney

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 7/20/25 09:58, Cursitor Doom wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard
<[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer

[Snip!...]

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge oscillator [...]

Wien! The guy's name was Wien, Max Wien.

Let's stamp out this Wein misspelling.

Jeroen Belleman

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Sun, 20 Jul 2025 12:58:49 -0400, "Edward Rawde"
<[email protected]d> wrote:

"Bill Sloman" <[email protected]> wrote in message news:105ihmc$3ahv8$[email protected]...

On 20/07/2025 5:58 pm, Cursitor Doom wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard
<[email protected]> wrote:

<snip>

For which I was originally responsible for (before BS and Edward took
it to a whole new level).

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge
oscillator's principle of operation and how they overcame this problem
to stabilise the loop in the most simple and elegant way.

Cursitor Doom has delusions of competence.

There are a variety of ways of stabilising a Wein Bridge, and none of them are all that elegant. What the thread did manage to
establish was that there were a couple of ways of getting the harmonic content of the nominally sinewave output close to 150dB
below the fundamental nanovolts in a notionally 1V peak-to-peak signal.

John May was the most influential contributor, mostly because he has actually built examples of that kind of circuit. Edward Rawde
was the most persistent.

I'm still persisting but I don't currently have time to find out why a circuit which stabilizes and produces low distortion in
LTSpice just hits the power rail when the same circuit is run in QSpice.

Be careful what you wonder about in this forum, Edward. Our
troll-in-chief Bill Sloman will use anything to inveigle you into yet
another of his pointless pissing contests.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Christopher Howard <[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer and have also been studying schematics for
some historical analog computers. An integrator is modeled with an
inverting op amp with a feedback capacitor.

In the circle test, a simple loop is constructed modeling the
differential equation y'' = -y. This is implemented with

Integrator A -> Integrator B -> inverter -> loop back to input of
Integrator A.

In ideal integration, the function y — which is the output of Integrator
B — would be a steady sinusoidal wave that does not increase or decrease
in amplitude. However, in practice the wave will either grow or decay in amplitude, once the integrators are started. I've found that, with
smaller capacity feedback capacitors, the wave tends to grow, and with
larger ones, it tends to decay. One can see this as a growing or
shrinking circle on an oscilloscope, by feeding the outputs from both integrators into an XY scope display.

Could somebody please explain why this happens? I'm not grasping the
basic cause of this. I've been trying to read up about op amp
(in)stability, like in amplifiers and voltage followers, but I'm not
seeing if/how there is a connection between that and what is going on
here.

In ideal world your circuit is marginally stable, just at the border
between dumping and infinitely growing ones. In real world, due to
non-ideal nature of components and due to tolerances you get circuit
that behaves slightly differently.

If you want useful result take circuit that is asymptotically stable
(that is damping) and provide it with some excitation, you should
see result close to theoretical one. Or, if you insist on
sinusoidal test add nonlinear amplitude limiter (as other suggest).

--
Waldek Hebisch

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 21/07/2025 7:25 am, Cursitor Doom wrote:

On Sun, 20 Jul 2025 12:58:49 -0400, "Edward Rawde"
<[email protected]d> wrote:

"Bill Sloman" <[email protected]> wrote in message news:105ihmc$3ahv8$[email protected]...

On 20/07/2025 5:58 pm, Cursitor Doom wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard
<[email protected]> wrote:

<snip>

Be careful what you wonder about in this forum, Edward. Our
troll-in-chief Bill Sloman will use anything to inveigle you into yet
another of his pointless pissing contests.

Cursitor Doom doesn't know enough to see the point in many of our
discussions. He's an anonymous troll, so having him complain that one of
his critics - who actually posts under his real name - is a troll is a
bit rich.

--
Bill Sloman, Sydney

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 21/07/2025 7:30 am, Waldek Hebisch wrote:

Christopher Howard <[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer and have also been studying schematics for
some historical analog computers. An integrator is modeled with an
inverting op amp with a feedback capacitor.

In the circle test, a simple loop is constructed modeling the
differential equation y'' = -y. This is implemented with

Integrator A -> Integrator B -> inverter -> loop back to input of
Integrator A.

In ideal integration, the function y — which is the output of Integrator >> B — would be a steady sinusoidal wave that does not increase or decrease >> in amplitude. However, in practice the wave will either grow or decay in
amplitude, once the integrators are started. I've found that, with
smaller capacity feedback capacitors, the wave tends to grow, and with
larger ones, it tends to decay. One can see this as a growing or
shrinking circle on an oscilloscope, by feeding the outputs from both
integrators into an XY scope display.

Could somebody please explain why this happens? I'm not grasping the
basic cause of this. I've been trying to read up about op amp
(in)stability, like in amplifiers and voltage followers, but I'm not
seeing if/how there is a connection between that and what is going on
here.

In ideal world your circuit is marginally stable, just at the border
between dumping and infinitely growing ones. In real world, due to
non-ideal nature of components and due to tolerances you get circuit
that behaves slightly differently.

If you want useful result take circuit that is asymptotically stable
(that is damping) and provide it with some excitation, you should
see result close to theoretical one. Or, if you insist on
sinusoidal test add nonlinear amplitude limiter (as other suggest).

A non-linear amplitude limiter is a clipper, and that adds a
predictable, if small harmonic components. The Wien Bridge (thanks to
Jeroen Belleman) is usually built with a more subtle form of amplitude
control which adds (or subtracts) a controllable amount of tolerably
linear gain to the circuit.

The classic version relied on a temperature dependent resistor, whose resistance changed as it carried more current. It had to have enough
thermal mass that the resistance didn't change much over a single
sinusoidal cycle. Modern versions rectify or demodulate the signal to
get an amplitude-dependent control signal that electronically controls
the gain in a way that leads to a stable output amplitude.

--
Bill Sloman, Sydney

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard <[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer and have also been studying schematics for
some historical analog computers. An integrator is modeled with an
inverting op amp with a feedback capacitor.

In the circle test, a simple loop is constructed modeling the
differential equation y'' = -y. This is implemented with

Integrator A -> Integrator B -> inverter -> loop back to input of
Integrator A.

In ideal integration, the function y � which is the output of Integrator
B � would be a steady sinusoidal wave that does not increase or decrease
in amplitude. However, in practice the wave will either grow or decay in >amplitude, once the integrators are started. I've found that, with
smaller capacity feedback capacitors, the wave tends to grow, and with
larger ones, it tends to decay. One can see this as a growing or
shrinking circle on an oscilloscope, by feeding the outputs from both >integrators into an XY scope display.

Could somebody please explain why this happens? I'm not grasping the
basic cause of this. I've been trying to read up about op amp
(in)stability, like in amplifiers and voltage followers, but I'm not
seeing if/how there is a connection between that and what is going on
here.

I was recently talking to my design center kids about diferential
equations and circuits. All three are CE/EE grads or students. I
mentioned "initial conditions" and one of them recalled hearing the
term. [1]

Your equation has an infinite number of solutions. One is Y=0, cold
and dead. One is a 1-volt sine wave. Another is a megavolt sine wave.
Which it does depends on the initial condition, what it's already
doing when you walk into the room. It will keep doing that. You can
Spice that.

In real life, there are no ideal integrators, so the loop has
additional phase shifts so there are different solutions to the
equation; specifically decaying or increasing sine waves.

To make a decent oscillator, you need an increasing amplitude circuit
and some sort of active amplitude limiter. Opamp clipping is the
cheapest limiter.


[1] Computer Engineering is a kinda oxymoron.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Jeroen Belleman <[email protected]> wrote:

On 7/20/25 09:58, Cursitor Doom wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard
<[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer

[Snip!...]

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge
oscillator [...]

Wien! The guy's name was Wien, Max Wien.

Let's stamp out this Wein misspelling.

Jeroen Belleman

I’ll take Riesling over schnitzel or torte any day. ;)

Cheers

Phil Hobbs

BTW Jeroen, seems like you’ve been getting out of bed on the wrong side lately.

--
Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Don't blame the poor opamps. They can't do y'' = -y perfectly. A
little extra phase shift always sneaks in.

At the peak frequency, if the loop gain is 1.001, the oscillation will
grow exponentially. If gain is 0.9999, it will decay exponentially.

In real life, there are no ideal integrators, so the loop has
additional phase shifts so there are different solutions to the
equation; specifically decaying or increasing sine waves.

Could you clarify: where is the phase shift happening and why? Is it in
the op amp, or the feedback capacitor, or...?

As mentioned, if I switch the integrator capacitor to a lower value —
which in turn increases the frequency of the y'' = -y model — then the oscillation amplitude grows quickly. But at the higher values (and lower frequency), it dampens instead.

The point of the circle test (y'' = -y), from what I've read and been
told, was to test how good an integrator was by seeing how fast it
deviated from the ideal circle. Each integrator output could be used for
X and Y, respectively, to generate the circle. But I haven't been able
to find an explanation on how you make the integrator good/better. The schematics I look at for various electronic analog integrator — in
historical analog computers — are all pretty much the same, except
sometimes different models of op amps are used.

I accept that, to have the perfectly stable wave, you must have some
kind of regulator. But it bothers me that, in my analog computer, the
deviation is very rapid. When using the high value capacitor, it damps
from IC to quarter wave in about four seconds. When using the low value capacitor, the wave stays reasonably stable for only about 50
milliseconds. I'm wondering how such dampening or growth is affecting
the accuracy of models where there is a lot of rapid oscillation going
on. E.g., spring-mass system like x'' = -(dx' + kx) / m. How do you know
how much deviation is going on? Or how do you minimize the impact of it
by building a better integrator?

--
Christopher Howard

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

"
On 21/07/2025 11:11 pm, john larkin wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard <[email protected]> wrote:

<snip>

I was recently talking to my design center kids about differential
equations and circuits. All three are CE/EE grads or students. I
mentioned "initial conditions" and one of them recalled hearing the
term. [1]

I got it drummed into my head during Pure Math 2, which was one of my
second year undergraduate courses, It was entirely about integral and differential calculus, and remarkably tedious, but useful.

Your equation has an infinite number of solutions. One is Y=0, cold
and dead. One is a 1-volt sine wave. Another is a megavolt sine wave.
Which it does depends on the initial condition, what it's already
doing when you walk into the room. It will keep doing that. You can
Spice that.

Since Spice is designed to evaluate the sort of differential equations
which described electronic circuit, this is a remarkably obvious assertion.

In real life, there are no ideal integrators, so the loop has
additional phase shifts, so there are different solutions to the
equation; specifically decaying or increasing sine waves.

The phase shifts control the frequency, not the amplitude, which is
determined by component ratios which drift as the ambient temperature
changes, or the components age.

To make a decent oscillator, you need an increasing amplitude circuit
and some sort of active amplitude limiter. Opamp clipping is the
cheapest limiter.

But it introduces harmonics of the sine wave being clipped.

[1] Computer Engineering is a kinda oxymoron.

Why? Computers are engineered devices, and designing them is definitely
a branch of engineering. It strikes me as a sub-branch of electrical engineering, or perhaps of just electronic engineering (which is itself
a sub-branch of electrical engineering).

An oxymoron embodies a contradiction. "Military intelligence" is the
classic example.

"According to the Oxford Learner's Dictionary, oxymoron is defined as “a phrase that combines two words that seem to be the opposite of each
other.” The Cambridge Dictionary defines an oxymoron as “two words or phrases used together that have, or seem to have, opposite meanings.”

--
Bill Sloman, Sydney

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 7/21/25 19:37, Christopher Howard wrote:

Don't blame the poor opamps. They can't do y'' = -y perfectly. A
little extra phase shift always sneaks in.

At the peak frequency, if the loop gain is 1.001, the oscillation will
grow exponentially. If gain is 0.9999, it will decay exponentially.

In real life, there are no ideal integrators, so the loop has
additional phase shifts so there are different solutions to the
equation; specifically decaying or increasing sine waves.

Could you clarify: where is the phase shift happening and why? Is it in
the op amp, or the feedback capacitor, or...?

As mentioned, if I switch the integrator capacitor to a lower value —
which in turn increases the frequency of the y'' = -y model — then the oscillation amplitude grows quickly. But at the higher values (and lower frequency), it dampens instead.

I wouldn't worry too much. Stable oscillation is really an edge case.
Your capacitors aren't likely to be better than +/- 5% or so, so
inevitably you are going to see the amplitude shrink or grow pretty
fast, a few percent per cycle.

I get the impression that you are mathematically inclined, and not
acquainted with the variability of real-life electronics.

In practical applications of analog computing, the amplitude would
probably grow or dampen *much* faster. As long as you aren't getting
near the conditions where the rate of change due to component
tolerances comes into play, your results should be near enough
correct to be useful.

Are you familiar with Laplace transforms and root-locus analysis?

Jeroen Belleman

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On 7/21/25 15:11, Phil Hobbs wrote:

Jeroen Belleman <[email protected]> wrote:

On 7/20/25 09:58, Cursitor Doom wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard
<[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I
built one analog computer

[Snip!...]

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge
oscillator [...]

Wien! The guy's name was Wien, Max Wien.

Let's stamp out this Wein misspelling.

Jeroen Belleman

I’ll take Riesling over schnitzel or torte any day. ;)

Cheers

Phil Hobbs

BTW Jeroen, seems like you’ve been getting out of bed on the wrong side lately.

You are right. I should work on contributing something useful, or keep
quiet. Sorry.

Max Wien invented the bridge that bears his name 'just' to measure
complex impedances. I believe it was Hewlett who first used it to
make an oscillator.

As has been amply stated in this thread, sine wave oscillators need
some mechanism to stabilize their amplitude, and that inevitably
introduces some distortion. There's a trade-off to be made between
the speed of the amplitude regulation and the distorion. We've had
a near-endless discussion about that in this group last year, but
as far as I recall, nobody actually built something to test the
theory and everyone is tired of the subject.

Jeroen Belleman

--- SoupGate-Win32 v1.05
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Bill Sloman <[email protected]> wrote:

[...]

"According to the Oxford Learner's Dictionary, oxymoron is defined as “a phrase that combines two words that seem to be the opposite of each
other

I always thought it meant a Stupid Cow.


--
~ Liz Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

--- SoupGate-Win32 v1.05
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Jeroen Belleman <[email protected]> wrote:

On 7/21/25 15:11, Phil Hobbs wrote:

Jeroen Belleman <[email protected]> wrote:

On 7/20/25 09:58, Cursitor Doom wrote:

On Sat, 19 Jul 2025 15:29:53 -0700, john larkin <[email protected]>
wrote:

On Sat, 19 Jul 2025 12:29:38 -0800, Christopher Howard
<[email protected]> wrote:

Hi, I'm continuing my exploration of electrical analog computing. I >>>>>> built one analog computer

[Snip!...]

You need to managed the gain actively to compensate for the effect
John mentioned. It might be instructive to look into the Wein Bridge
oscillator [...]

Wien! The guy's name was Wien, Max Wien.

Let's stamp out this Wein misspelling.

Jeroen Belleman

I’ll take Riesling over schnitzel or torte any day. ;)

Cheers

Phil Hobbs

BTW Jeroen, seems like you’ve been getting out of bed on the wrong side
lately.

You are right. I should work on contributing something useful, or keep
quiet. Sorry.

No worries—I’ve had to repent for my bad temper in these hallowed halls on occasion. ;)

Max Wien invented the bridge that bears his name 'just' to measure
complex impedances. I believe it was Hewlett who first used it to
make an oscillator.

As has been amply stated in this thread, sine wave oscillators need
some mechanism to stabilize their amplitude, and that inevitably
introduces some distortion. There's a trade-off to be made between
the speed of the amplitude regulation and the distorion. We've had
a near-endless discussion about that in this group last year, but
as far as I recall, nobody actually built something to test the
theory and everyone is tired of the subject.

Jeroen Belleman

Yup. Five iterations of Newton-Raphson will make an oscillator
amplitude-stable in LTspice to machine accuracy.

Cheers

Phil Hobbs




--
Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics

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I wouldn't worry too much. Stable oscillation is really an edge case.
Your capacitors aren't likely to be better than +/- 5% or so, so
inevitably you are going to see the amplitude shrink or grow pretty
fast, a few percent per cycle.

I'm not quite following this. I'm not currently tuning things that finely in my analog computer, but I could. I think the normal approach with analog computers in the past was (1) component tolerance of 1% or better; and (2) insert a gain adjustment knob
right before the capacitor(s) so you can speed or slow down the integration to a precise time unit, using a counter. Or, alternatively, you could use a comparator and test two integrators at a time just to make sure they are all at the same time unit.

I used high precision resistors — I think they were 0.25%. I'm not sure though what the tolerance is on these polyester capacitors that I had handy. A batch I got off Amazon a few years back. 10nF is 2A103J.

Are you familiar with Laplace transforms and root-locus analysis?

I'm still early on in my study of differential equations, and Laplace transforms aren't covered until one of the late chapters of the textbook I am using. I don't think I studied root-locus analysis yet.

--
Christopher Howard

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On 7/21/25 18:36, Christopher Howard wrote:

I wouldn't worry too much. Stable oscillation is really an edge
case. Your capacitors aren't likely to be better than +/- 5% or
so, so inevitably you are going to see the amplitude shrink or
grow pretty fast, a few percent per cycle.

I'm not quite following this. I'm not currently tuning things that
finely in my analog computer, but I could. I think the normal
approach with analog computers in the past was (1) component
tolerance of 1% or better; and (2) insert a gain adjustment knob
right before the capacitor(s) so you can speed or slow down the
integration to a precise time unit, using a counter. Or,
alternatively, you could use a comparator and test two integrators
at a time just to make sure they are all at the same time unit.

I used high precision resistors — I think they were 0.25%. I'm not
sure though what the tolerance is on these polyester capacitors that
I had handy. A batch I got off Amazon a few years back. 10nF is
2A103J.

J tolerance is +-5%. K is +-10%, M is +-20%. Lots worse than the
resistors.

Are you familiar with Laplace transforms and root-locus analysis? Single-sided Laplace and root locus are pretty well obsolete at this

point, even for folks like yours truly, who has done a lot of hand calculations. (Root locus is sometimes useful in complicated control
systems, but that's mostly for control system gurus like our much-missed
former regular Tim Wescott.)

Single-sided Laplace really applies only to initial-value problems.
Those are nice for teaching purposes but have only a very oblique
connection to the real world, where there is all sorts of external
forcing after power-up, due to supply voltage, load variations,
temperature, circuit noise, yada yada.

Loop stability can be broken up into small-signal stability, where the
gain doesn't vary with signal level, and large-signal stability, where
it does. Small-signal stability is easiest to understand in the
frequency domain, using frequency compensation ideas. (The difference
between the two is responsible for oscillator startup problems, which is
also an interesting topic, but at the moment we're trying not to build
an oscillator.)

As indicated, small-signal stability lives in the Fourier domain, or the two-sided Laplace domain, which is exactly the same except with the plot rotated 90 degrees. The one-sided Laplace is derivable from them. (The
Z transform is also trivially related to them, which makes discrete-time processes more intelligible to circuits folk, but I digress.)

A feedback system's stability is governed by its _phase margin_, which
is the difference between the closed-loop phase shift and 360 degrees, evaluated at the frequency where the loop gain drops to 1.0.

(Deep breaths) Some context might possibly be helpful. ;)

An amplitude-stable electronic oscillator is a feedback loop whose
round-trip gain is 1.0, with zero phase, at some frequency f_0. That is,
first you break the feedback loop at some convenient point, making one
side the input and the other side the output. (If it isn't obvious which
side is which, split the loop somewhere else.)

Then, if you drive the input side with

V_in = A sin (2 pi f_0 t),

the output will produce exactly the same waveform. One has to correct
for source and load impedance, as well as possible nonlinearity, but
this is the gist--stability is basically a linear problem.

The complex-valued ratio of the output signal to the input is called the _loop_gain_. (If you're not comfortable with complex notation for
circuits, we can talk about that too.)

In order for the oscillation to be stable, the magnitude of the loop
gain |A_VL| has to be exactly unity. Loosely speaking, if the initial
signal is 1, the first pass round the loop will make it A_VL, the second
pass A_VL**2, and so on. So if |A_VL| isn't 1.0000..., you don't get
amplitude stability.

In the same vein, the phase of the loop gain has to be an integer
multiple of 2 pi, or else the phase will keep walking, i.e. f_0 isn't
what we thought it was.

Feedback loops also need to be stable at DC. You can't phase-shift DC,
so you have two choices: inverting (phi = 180 degrees) or noninverting
(phi = 0). For a stable loop, we have to pick the inverting case, so at DC,

arg{A_VL} = 180 degrees.

Thus we usually think about the _loop phase_ as ranging from 0 to -180
degrees, i.e. the range of additional delays between an inverting DC amp
and an oscillator. (The electrical engineering sign convention is that
a delay has a negative phase.)

If |A_VL| < 1 when the loop phase reaches -180 degrees, the oscillation condition is not met, and the loop is stable. (Though not necessarily well-behaved--see below.) The ratio 1/|A_VL| at 180 degrees is called
the _gain margin_ of the loop, and is a measure of its stability.

Equivalently, if the loop phase arg(A_VL) hasn't reached -180 degrees at
the frequency f_0 where |A_VL| = 1.0, the loop is stable. The
_phase_margin_

phi_M = 180 degrees + arg(A_VL) @ f_0

is also a measure of loop stability. (Remember, arg(A_VL) < 0.)

Of the two, phase margin is more useful in design work.

A nice critically-damped second-order system has a phase margin of about
65 degrees, iirc. At higher values the system is overdamped, and at
lower values it's underdamped, so that the impulse response exhibits
overshoot and rings. The ringing gets worse quite rapidly as phi_M
declines, and for most purposes 45 degrees is effectively the limit.
This is more or less true even for higher-order loops--65 degrees gets
you a nice monotonic step response, anything less will ring at least a
bit. As phi_M goes to 0, the ringing gets worse and worse, and when
phi_M reaches zero, it becomes a continuous oscillation.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

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On Mon, 21 Jul 2025 09:37:14 -0800, Christopher Howard <[email protected]> wrote:

Don't blame the poor opamps. They can't do y'' = -y perfectly. A
little extra phase shift always sneaks in.

At the peak frequency, if the loop gain is 1.001, the oscillation will
grow exponentially. If gain is 0.9999, it will decay exponentially.

In real life, there are no ideal integrators, so the loop has
additional phase shifts so there are different solutions to the
equation; specifically decaying or increasing sine waves.

Could you clarify: where is the phase shift happening and why? Is it in
the op amp, or the feedback capacitor, or...?

As mentioned, if I switch the integrator capacitor to a lower value �
which in turn increases the frequency of the y'' = -y model � then the >oscillation amplitude grows quickly. But at the higher values (and lower >frequency), it dampens instead.

The point of the circle test (y'' = -y), from what I've read and been
told, was to test how good an integrator was by seeing how fast it
deviated from the ideal circle. Each integrator output could be used for
X and Y, respectively, to generate the circle. But I haven't been able
to find an explanation on how you make the integrator good/better. The >schematics I look at for various electronic analog integrator � in
historical analog computers � are all pretty much the same, except
sometimes different models of op amps are used.

I accept that, to have the perfectly stable wave, you must have some
kind of regulator. But it bothers me that, in my analog computer, the >deviation is very rapid. When using the high value capacitor, it damps
from IC to quarter wave in about four seconds. When using the low value >capacitor, the wave stays reasonably stable for only about 50
milliseconds. I'm wondering how such dampening or growth is affecting
the accuracy of models where there is a lot of rapid oscillation going
on. E.g., spring-mass system like x'' = -(dx' + kx) / m. How do you know
how much deviation is going on? Or how do you minimize the impact of it
by building a better integrator?

Two serious limits are the finite gain-bandwidth of the opamps and
losses in capacitors.

I'd Spice it to get a feel for things. I like to use Universal Opamp2
so I can fiddle the gain-bandwidth. And one can add series and
parallel resistance to the caps.

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On 22/07/2025 5:28 am, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...]

"According to the Oxford Learner's Dictionary, oxymoron is defined as “a
phrase that combines two words that seem to be the opposite of each
other

I always thought it meant a Stupid Cow.

An ox is male. A cow is female. It may be fashionable to ignore gender differences, but it is imprecise.

--
Bill Sloman, Sydney

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On 22/07/2025 1:11 pm, john larkin wrote:

On Mon, 21 Jul 2025 09:37:14 -0800, Christopher Howard <[email protected]> wrote:

<snip>

Two serious limits are the finite gain-bandwidth of the opamps and
losses in capacitors.

I'd Spice it to get a feel for things. I like to use Universal Opamp2
so I can fiddle the gain-bandwidth. And one can add series and
parallel resistance to the caps.

If you use Spice without putting in the data sheet equivalents series
and parallel resistances and inductances for you capacitors you aren't
using it to simulate the real circuit you are trying to simulate.

Most resistor have a about 0.05pF of parallel capacitance which rarely
matter (but if you are using 1M resistors at high frequencies, it does).

Inductors almost always have significant parallel capacitance and series resistance, and mostly both matter.

--
Bill Sloman, Sydney

--- SoupGate-Win32 v1.05
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Bill Sloman <[email protected]> wrote:

On 22/07/2025 5:28 am, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...]

"According to the Oxford Learner's Dictionary, oxymoron is defined as
�"a phrase that combines two words that seem to be the opposite of
each other

I always thought it meant a Stupid Cow.

An ox is male. A cow is female. It may be fashionable to ignore gender differences, but it is imprecise.

Oxen are a type of bovine, bull and cow are the sexes.


--
~ Liz Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 22/07/2025 6:21 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

On 22/07/2025 5:28 am, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...]

"According to the Oxford Learner's Dictionary, oxymoron is defined as
â€Å"a phrase that combines two words that seem to be the opposite of
each other

I always thought it meant a Stupid Cow.

An ox is male. A cow is female. It may be fashionable to ignore gender
differences, but it is imprecise.

Oxen are a type of bovine, bull and cow are the sexes.

Oxen are usually castrated males. They are always trained daft animals,
and the name is reserved for mature animals over four years of age.
Again, you are being imprecise. This undercuts any comic effect you
might have been aiming for.

--
Bill Sloman, Sydney

--- SoupGate-Win32 v1.05
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Thanks for the response Phil, seems like the most helpful one so far.

Single-sided Laplace really applies only to initial-value problems.
Those are nice for teaching purposes but have only a very oblique
connection to the real world, where there is all sorts of external
forcing after power-up, due to supply voltage, load variations,
temperature, circuit noise, yada yada.

Sorry, I got lost on the discussion of Laplace and Z-transform and such
like. Hopefully I'll catch up with you in another few months of study.

The complex-valued ratio of the output signal to the input is called
the _loop_gain_. (If you're not comfortable with complex notation for circuits, we can talk about that too.)

I'm familiar with the basic idea of complex numbers and relation to
circuits (i.e., inductance/capacitance as imaginary) but I'm not
practiced in circuit analysis and more hand-holding would be appreciated.
I explored complex numbers a little once a few years ago in some studies
in SDR DSP, in a ham radio context.

In order for the oscillation to be stable, the magnitude of the loop
gain |A_VL| has to be exactly unity. Loosely speaking, if the initial
signal is 1, the first pass round the loop will make it A_VL, the
second pass A_VL**2, and so on. So if |A_VL| isn't 1.0000..., you
don't get amplitude stability.

In the same vein, the phase of the loop gain has to be an integer
multiple of 2 pi, or else the phase will keep walking, i.e. f_0 isn't
what we thought it was.

...snip...

A nice critically-damped second-order system has a phase margin of
about 65 degrees, iirc. At higher values the system is overdamped,
and at lower values it's underdamped, so that the impulse response
exhibits overshoot and rings. The ringing gets worse quite rapidly as
phi_M declines, and for most purposes 45 degrees is effectively the
limit. This is more or less true even for higher-order loops--65
degrees gets you a nice monotonic step response, anything less will
ring at least a bit. As phi_M goes to 0, the ringing gets worse and
worse, and when phi_M reaches zero, it becomes a continuous
oscillation.

Help me as I try to process this: so, a particular amp op model has
a certain phase margin for a given frequency, which is the "safety
margin" (wikipedia) where the output won't at least grow in amplitude.
Now assume that I'm just using op amps as integrators with feedback
capacitors and input resistors, and that the outputs are just connected
to other integrators of like design (not some really weird load): can I calculate my phase just by looking at the input resistor, feedback
capacitor, and a target frequency?

In my application, aiming for accurate modeling of, say, a spring
damper system, I would think the goal would be to stay as close as
possible to the phase where I would have unity gain. So maybe I would
need to speed or slow down the simulation with a coefficient variable
resistor, to stay near an ideal frequency?

I had trouble following the part where you explained how we stay stable
at DC, though I'm not doubting it was a good explanation. I'll try to study/review this some more.

--
Christopher Howard

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On Tue, 22 Jul 2025 10:02:36 -0800, Christopher Howard <[email protected]> wrote:

Thanks for the response Phil, seems like the most helpful one so far.

Single-sided Laplace really applies only to initial-value problems.
Those are nice for teaching purposes but have only a very oblique
connection to the real world, where there is all sorts of external
forcing after power-up, due to supply voltage, load variations,
temperature, circuit noise, yada yada.

Sorry, I got lost on the discussion of Laplace and Z-transform and such
like. Hopefully I'll catch up with you in another few months of study.

The complex-valued ratio of the output signal to the input is called
the _loop_gain_. (If you're not comfortable with complex notation for
circuits, we can talk about that too.)

I'm familiar with the basic idea of complex numbers and relation to
circuits (i.e., inductance/capacitance as imaginary) but I'm not
practiced in circuit analysis and more hand-holding would be appreciated.
I explored complex numbers a little once a few years ago in some studies
in SDR DSP, in a ham radio context.

In order for the oscillation to be stable, the magnitude of the loop
gain |A_VL| has to be exactly unity. Loosely speaking, if the initial
signal is 1, the first pass round the loop will make it A_VL, the
second pass A_VL**2, and so on. So if |A_VL| isn't 1.0000..., you
don't get amplitude stability.

In the same vein, the phase of the loop gain has to be an integer
multiple of 2 pi, or else the phase will keep walking, i.e. f_0 isn't
what we thought it was.

...snip...

A nice critically-damped second-order system has a phase margin of
about 65 degrees, iirc. At higher values the system is overdamped,
and at lower values it's underdamped, so that the impulse response
exhibits overshoot and rings. The ringing gets worse quite rapidly as
phi_M declines, and for most purposes 45 degrees is effectively the
limit. This is more or less true even for higher-order loops--65
degrees gets you a nice monotonic step response, anything less will
ring at least a bit. As phi_M goes to 0, the ringing gets worse and
worse, and when phi_M reaches zero, it becomes a continuous
oscillation.

Help me as I try to process this: so, a particular amp op model has
a certain phase margin for a given frequency, which is the "safety
margin" (wikipedia) where the output won't at least grow in amplitude.
Now assume that I'm just using op amps as integrators with feedback >capacitors and input resistors, and that the outputs are just connected
to other integrators of like design (not some really weird load): can I >calculate my phase just by looking at the input resistor, feedback
capacitor, and a target frequency?

In my application, aiming for accurate modeling of, say, a spring
damper system, I would think the goal would be to stay as close as
possible to the phase where I would have unity gain. So maybe I would
need to speed or slow down the simulation with a coefficient variable >resistor, to stay near an ideal frequency?

I had trouble following the part where you explained how we stay stable
at DC, though I'm not doubting it was a good explanation. I'll try to >study/review this some more.

I good way to build physical instinct is to play with a microphone on
a HiFi system in a room. As you turn the volume up, you will get to a
point where the room seems to ring a bit - this where your phase
margin is marginal. If you turn the volume up even more, it will howl
or screech. The loop gain at some frequency now exceeds unity.

Joe

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On 23/07/2025 4:02 am, Christopher Howard wrote:

Thanks for the response Phil, seems like the most helpful one so far.

<snip>

Help me as I try to process this: so, a particular amp op model has
a certain phase margin for a given frequency, which is the "safety
margin" (wikipedia) where the output won't at least grow in amplitude.
Now assume that I'm just using op amps as integrators with feedback capacitors and input resistors, and that the outputs are just connected
to other integrators of like design (not some really weird load): can I calculate my phase just by looking at the input resistor, feedback
capacitor, and a target frequency?

Op amps intended to be stabilised by negative feedback have have a
particular variation of phase angle and gain with frequency (Bode plot)
where the phase shift doesn't reach 180 degrees below a frequency lower
than the one where the gain has dropped to less than one.

Easy to use op amps have a nice smooth linear rise in phase shift with frequency with an equally smooth decline in gain. Crankier op amps can
still be useful.

You need to be aware of the Bode plot of the op amp you are planning to
use, as well as the input resistor, the feedback capacitor and your
target frequency. If the op amp hasn't got enough bandwidth, the circuit
won't work.

In my application, aiming for accurate modeling of, say, a spring
damper system, I would think the goal would be to stay as close as
possible to the phase where I would have unity gain. So maybe I would
need to speed or slow down the simulation with a coefficient variable resistor, to stay near an ideal frequency?

You'd need to sped up or slow down the simulated circuit. The simulation
is just modelling the circuit you have chosen to simulate.

I had trouble following the part where you explained how we stay stable
at DC, though I'm not doubting it was a good explanation. I'll try to study/review this some more.

Lot's of people have trouble working out how a Wien bridge can be
engineered to run at a constant stable amplitude. My opinion is that
this always requires adding some kind of non-linear element to control
the gain around the loop, and if you want a low harmonic content in the
output you generate, and a precise output level, this has include a
precision rectifier or demodulator whose output can be compared with a
stable reference voltage to generate the gain-adjusting signal.

Less picky individuals tend to cut corners, and so have I when the
application is less demanding.

--
Bill Slomanh, sydney

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Bill Sloman <[email protected]> wrote:

On 22/07/2025 6:21 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

On 22/07/2025 5:28 am, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...] > "According to the Oxford Learner's Dictionary, oxymoron is >>>defined as > â€Å"a phrase that combines two words that seem >>>to be the opposite of > each other

I always thought it meant a Stupid Cow.

An ox is male. A cow is female. It may be fashionable to ignore gender
differences, but it is imprecise.

Oxen are a type of bovine, bull and cow are the sexes.

Oxen are usually castrated males.

Some may be castrated males but, according to my dictionary they are
used for draught, for milk or for meat. Unless you are going to tell me
you have succesfully milked a bull, "Stupid Cow" makes sense.


--
~ Liz Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

--- SoupGate-Win32 v1.05
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On 23/07/2025 7:20 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

On 22/07/2025 6:21 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

On 22/07/2025 5:28 am, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...] > "According to the Oxford Learner's Dictionary, oxymoron is
defined as > â€Å"a phrase that combines two words that seem
to be the opposite of > each other

I always thought it meant a Stupid Cow.

An ox is male. A cow is female. It may be fashionable to ignore gender >>>> differences, but it is imprecise.

Oxen are a type of bovine, bull and cow are the sexes.

Oxen are usually castrated males.

Some may be castrated males but, according to my dictionary they are
used for draft, for milk or for meat.

The dictionaries I've consulted concentrated on the draft animal aspect. Ignorant people do use words on the basis of their idiosyncratic
understanding of what they mean, but the intelligent ignorant do
generally learn to do better.

Unless you are going to tell me
you have succesfully milked a bull, "Stupid Cow" makes sense.

Since I've never been in the bovine artificial insemination business, I
can't ever claim to have "milked" a bull.

This has nothing to do with your bizarre equation of "stupid cow" which
is purely English, with "oxymoron" which is comes to us from Greek.
There is a word "boustrophedron" meaning

"written from right to left and from left to right in alternate lines."

which also comes to us from Greek, which does involve oxen, since the
text is written onto the page in the same way that an ox plows a field,
with the direction reversing every time the stream of text hits a page boundary.

It's a compound of "bous" - ox - and "stophos" - turning.

Electron beam microfabricators write their patterns in the same way, and
for much the same reason - it takes time to get the beam/ox back to the
other side of the field being written/plowed.

--
Bill Sloman, Sydney

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Bill Sloman <[email protected]> wrote:

[...]

Oxen are usually castrated males.

Some may be castrated males but, according to my dictionary they are
used for draft, for milk or for meat.

The dictionaries I've consulted concentrated on the draft animal aspect. Ignorant people do use words on the basis of their idiosyncratic understanding of what they mean, but the intelligent ignorant do
generally learn to do better.

I am amazed that I managed to work in animal husbandry for 30 years and
obtain a degree in biology without your helpful advice to enlighten my ignorance.

Unless you are going to tell me
you have succesfully milked a bull, "Stupid Cow" makes sense.

Since I've never been in the bovine artificial insemination business, I
can't ever claim to have "milked" a bull.

Well I have been in that business and I can assure you that the 'milk'
you get from a bull is not something you would want to waste on your cornflakes, there are more productive places to put it (and it is very expensive to buy, if the bull is of good repute). I have no reason to
doubt that the milk referred to in the dictionary confirms the common
existence of oxen cows.



--
~ Liz Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

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Bill Sloman <[email protected]> wrote:

On 23/07/2025 10:41 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...]

Oxen are usually castrated males.

Some may be castrated males but, according to my dictionary they are
used for draft, for milk or for meat.

The dictionaries I've consulted concentrated on the draft animal aspect. >> Ignorant people do use words on the basis of their idiosyncratic
understanding of what they mean, but the intelligent ignorant do
generally learn to do better.

I am amazed that I managed to work in animal husbandry for 30 years and obtain a degree in biology without your helpful advice to enlighten my ignorance.

John Larkin has worked in electronics for just as long and still seems
to need advice (not that he appreciates it when he gets it).

[...]

Puns don't make any sense at all, and the proper response to any pun is always derision.

Perhaps you should ignore my puns in future and I, in return, shall
ignore your advice on non-electronic subjects.


--
~ Liz Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

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On 24/07/2025 12:45 am, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

On 23/07/2025 10:41 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

<snip>

It's a cross-linguistic pun on oxymoron, and has no other justification.
Puns don't make any sense at all, and the proper response to any pun is always derision.

Perhaps you should ignore my puns in future

Puns always earn derision.

and I, in return, shall ignore your advice on non-electronic subjects.

That's no kind of a fair exchange. I'm perfectly used to having my
advice ignored on a whole range of subjects.

--
Bill Sloman, Sydney

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On 23/07/2025 10:41 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...]

Oxen are usually castrated males.

Some may be castrated males but, according to my dictionary they are
used for draft, for milk or for meat.

The dictionaries I've consulted concentrated on the draft animal aspect.
Ignorant people do use words on the basis of their idiosyncratic
understanding of what they mean, but the intelligent ignorant do
generally learn to do better.

I am amazed that I managed to work in animal husbandry for 30 years and obtain a degree in biology without your helpful advice to enlighten my ignorance.

John Larkin has worked in electronics for just as long and still seems
to need advice (not that he appreciates it when he gets it).

Unless you are going to tell me
you have succesfully milked a bull, "Stupid Cow" makes sense.

It's a cross-linguistic pun on oxymoron, and has no other justification.
Puns don't make any sense at all, and the proper response to any pun is
always derision.

Since I've never been in the bovine artificial insemination business, I
can't ever claim to have "milked" a bull.

Well I have been in that business and I can assure you that the 'milk'
you get from a bull is not something you would want to waste on your cornflakes, there are more productive places to put it (and it is very expensive to buy, if the bull is of good repute).

That I have heard. The boarding school I got sent to in Tasmania for the
last fours of my secondary education also looked after a lot of farmer's
sons. My parents both had university degrees in chemistry, and chemistry
didn't generate nearly as much discussion with my schoolmates.

I have no reason to
doubt that the milk referred to in the dictionary confirms the common existence of oxen cows.

A dictionary isn't a great place to find out the sex ratio of draft
animals. They want to find examples where a word has been used - old
examples are interesting, numerous examples much less so.

http://webhome.auburn.edu/~nunnath/engl6240/kucera67.html

gives the first 2200 words in the Kucera-Francis table of word
frequencies in American English. It dates from 1967. Dictionaries go
back rather further.

--
Bill Sloman, Sydney

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On Wed, 23 Jul 2025 15:45:56 +0100, [email protected]d
(Liz Tuddenham) wrote:

Bill Sloman <[email protected]> wrote:

On 23/07/2025 10:41 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...]

Oxen are usually castrated males.

Some may be castrated males but, according to my dictionary they are
used for draft, for milk or for meat.

The dictionaries I've consulted concentrated on the draft animal aspect. >> >> Ignorant people do use words on the basis of their idiosyncratic
understanding of what they mean, but the intelligent ignorant do
generally learn to do better.

I am amazed that I managed to work in animal husbandry for 30 years and
obtain a degree in biology without your helpful advice to enlighten my
ignorance.

John Larkin has worked in electronics for just as long and still seems
to need advice (not that he appreciates it when he gets it).

[...]

Puns don't make any sense at all, and the proper response to any pun is
always derision.

Perhaps you should ignore my puns in future and I, in return, shall
ignore your advice on non-electronic subjects.

Why limit yourself?

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Bill Sloman <[email protected]> wrote:

On 24/07/2025 12:45 am, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

On 23/07/2025 10:41 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

<snip> >> It's a cross-linguistic pun on oxymoron, and has no other justification. >> Puns don't make any sense at all, and the proper
response to any pun is always derision. >

Perhaps you should ignore my puns in future

Puns always earn derision.

...but you don't always have to post your every thought, especially the
copious negative ones.

and I, in return, shall ignore your advice on non-electronic subjects.

That's no kind of a fair exchange. I'm perfectly used to having my
advice ignored on a whole range of subjects.

It sounds as though I should expand the range of subjects to make it
fairer.


--
~ Liz Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

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On 24/07/2025 2:13 am, john larkin wrote:

On Wed, 23 Jul 2025 15:45:56 +0100, [email protected]d
(Liz Tuddenham) wrote:

Bill Sloman <[email protected]> wrote:

On 23/07/2025 10:41 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...]

Oxen are usually castrated males.

Some may be castrated males but, according to my dictionary they are >>>>>> used for draft, for milk or for meat.

The dictionaries I've consulted concentrated on the draft animal aspect. >>>>> Ignorant people do use words on the basis of their idiosyncratic
understanding of what they mean, but the intelligent ignorant do
generally learn to do better.

I am amazed that I managed to work in animal husbandry for 30 years and >>>> obtain a degree in biology without your helpful advice to enlighten my >>>> ignorance.

John Larkin has worked in electronics for just as long and still seems
to need advice (not that he appreciates it when he gets it).

[...]

Puns don't make any sense at all, and the proper response to any pun is
always derision.

Perhaps you should ignore my puns in future and I, in return, shall
ignore your advice on non-electronic subjects.

Why limit yourself?

She could become a second Donald Trump. The UK seems to be less
well-supplied with gullible twits as the US, so it wouldn't have the
career prospects of the American version, but Liz Truss might be a
UK-specific role model. Liz Truss does seem to have ignored a great deal
of good advice - not as much as Donald Trump, but enough to command
quite a lot of well-deserved derision.

As the book "The Big Myth" demonstrates, US big business has been
working on maximising the gullibility of American voters for decades,
and it has paid off - if not a way that will help US big business.
Protecting them from foreign competition isn't going to make them more efficient or more productive. Protection against foreign competition may
be a way to get local industry going from scratch, but if you've
off-shored all your manufacturing, tariff protection is less helpful.

https://en.wikipedia.org/wiki/The_Big_Myth

--
Bill Sloman, Sydney

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On 24/07/2025 2:34 am, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

On 24/07/2025 12:45 am, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

On 23/07/2025 10:41 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

<snip> >> It's a cross-linguistic pun on oxymoron, and has no other
justification. >> Puns don't make any sense at all, and the proper
response to any pun is always derision. >

Perhaps you should ignore my puns in future

Puns always earn derision.

...but you don't always have to post your every thought, especially the copious negative ones.

Believe you me, I don't.

and I, in return, shall ignore your advice on non-electronic subjects.

That's no kind of a fair exchange. I'm perfectly used to having my
advice ignored on a whole range of subjects.

It sounds as though I should expand the range of subjects to make it
fairer.

That's one way of thinking about it. I'd prefer you raised your game so
that you gave us less to complain about.

--
Bill Sloman, Sydeney

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Bill Sloman <[email protected]> wrote:

[...]

That's one way of thinking about it. I'd prefer you raised your game so
that you gave us less to complain about.

Is that the royal "us" (like the royal "we") or has someone else
complained?


--
~ Liz Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

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On 24/07/2025 4:50 pm, Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...]

That's one way of thinking about it. I'd prefer you raised your game so
that you gave us less to complain about.

Is that the royal "us" (like the royal "we") or has someone else
complained?

More the collective us. As a group, we do a lot of complaining.

--
Bill Sloman, Sydney

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On 2025-07-23 11:50 p.m., Liz Tuddenham wrote:

Bill Sloman <[email protected]> wrote:

[...]

That's one way of thinking about it. I'd prefer you raised your game so
that you gave us less to complain about.

Is that the royal "us" (like the royal "we") or has someone else
complained?

Bill doesn't speak for me, as I (for one) appreciate your work and notes
here.

John :-#)#

--
(Please post followups or tech inquiries to the USENET newsgroup)
John's Jukes Ltd.
#7 - 3979 Marine Way, Burnaby, BC, Canada V5J 5E3
(604)872-5757 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."

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